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Adiabatic/Reversible application of external field to Ideal Gas

  1. Mar 7, 2009 #1

    a_h

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    Hello Fellow Physicists!

    I'm having trouble with a problem from a graduate course in Statistical Mechanics. It's a two part question; I've got the first part, it's the second part that's giving me trouble.

    The problem is about reversible/adiabatic processes, thinking of "reversible" as meaning "constant entropy." In part a (which I got, but it may help to have it here), we need to show that the evolution of a classical ideal gas obeys

    [tex]
    TV^{\frac{2}{3}}=\mbox{constant.}
    [/tex]

    I did this by using [tex] S=-\frac{\partial F}{\partial T} [/tex] with [tex] F=U-\mu N, U=\frac{3}{2}NT [/tex]. To get [tex] \frac{\partial \mu}{\partial T} [/tex], I solved

    [tex]
    e^{\frac{\mu}{T}}=n\lambda^{3}=\frac{N}{V} \frac{h^{3}}{(2\pi mT)^{3/2}}
    [/tex]

    for mu. ([tex] \lambda [/tex] is the thermodynamic wavelength, n is the number density of the gas.)

    Now for part b. We have a classical ideal gas in a cylinder. It is in equilibrium at a temperature [tex] T_{0} [/tex]. Now we slowly (adiabatically and reversibly) apply a uniform external field to the container, in a direction along its axis of symmetry. When we are done, there is a potential in our container given by

    [tex]
    u(z)=fz
    [/tex]

    where z is the distance along that symmetry axis and f is a constant. The question is, what is the temperature of the gas at the end of the process?

    I would think that we should use the same assumptions:

    [tex] S=-\frac{\partial F}{\partial T}=\mbox{constant} [/tex]

    and [tex] F=U-\mu N [/tex], but I don't know how u(z) gets incorporated into U. Any ideas?

    Thanks for all of your time, everyone.
     
  2. jcsd
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